Respuesta :
Answer:
Time period = 5.03 s
Explanation:
Given:
Maximum speed of object in SHM is, [tex]v_{max}=4.00[/tex] m/s
Maximum acceleration of object in SHM is, [tex]a_{max}=5.00[/tex] m/s².
Mass of object is 3.00 kg.
We know that for an object in SHM, the velocity is given as:
[tex]v=-\omega A \sin(\omega t)[/tex]
Where,
[tex]v_{max}=\omega A\\\omega \rightarrow \textrm{angular frequency}\\A\rightarrow \textrm{Amplitude}\\t\rightarrow time[/tex]
Therefore, [tex]\omega A=4.00[/tex] ------- 1
Also, for an object in SHM, the acceleration is given as:
[tex]v=-\omega^2 A \cos(\omega t)[/tex]
Where, [tex]a_{max}=\omega^2 A=5.00[/tex] ---- 2
Now, dividing equation 2 by 1, we get
[tex]\frac{\omega^2 A}{\omega A}=\frac{5.00}{4.00}\\\omega=1.25[/tex]
Now, time period is given as:
[tex]T=\frac{2\pi}{\omega}\\T=\frac{2\pi}{1.25}=5.03\textrm{ s}[/tex]
Therefore, the time period of simple harmonic motion is 5.03 s.
Period of harmonic motion is the time taken by it to complete one cycle of oscillatory motion.
The period of simple harmonic motion is 5.03s.
What is simple harmonic motion?
Period of simple harmonic motion is the time taken by it to complete one cycle of oscillatory motion.
The period of the function in terms of angular frequency can be given as,
[tex]T=\dfrac{2\pi}{\omega}[/tex]
Given information-
The mass of the object is 3 kg.
The maximum speed of the object is 4.00 m/s.
The maximum acceleration of the object is 5.00 m/s2.
The acceleration of the simple harmonic motion can be given as,
[tex]a_{max}=\omega^2 A\\5=(\omega A)\omega[/tex] .... 1
As the velocity of the simple harmonic motion can be given as,
[tex]v=\omega A[/tex]
Thus put the valued in the above equation 1 as,
[tex]5=(v)\omega\\5=4\omega\\\omega=\dfrac{5}{4} \\\omega=1.25[/tex]
Using the formula of period of simple harmonic motion,
[tex]T=\dfrac{2\pi}{\omega}\\ T=\dfrac{2\pi}{1.25}\\T=5.03 s[/tex]
Hence the period of simple harmonic motion is 5.03s.
Learn more about the simple harmonic motion here;
https://brainly.com/question/17315536
